0
$\begingroup$

Let $B\subset \mathbb R^2$ be given as a unit ball. Let $\omega\subset W^{2,2}(B)\cap L^\infty$ be given. (So we are only in 2d space)

Let $\Gamma\subset B$ be a closed Lipschitz curve such that $\mathcal H^1(\Gamma)<\infty$. Let $\Gamma_n\subset B$ be a sequence of closed Lipschitz curve such that $$ d_\mathcal H(\Gamma,\Gamma_n)\to 0 $$ and $$ \limsup_{n\to\infty}\mathcal H^1(\Gamma_n)<\infty. $$

Let function $u$ and $u_n$ be the solution such that $$ \begin{cases} -\Delta u=0&x\in\Omega\setminus \Gamma\\ u=\omega&x\in\partial B \end{cases} \text{ and } \begin{cases} -\Delta u_n=0&x\in\Omega\setminus \Gamma_n\\ u_n=\omega&x\in\partial B \end{cases} $$ So the boundary value on curve is free.

My question: can we have $$ \int_B|\nabla u_n|^2dx\to \int_B|\nabla u|^2dx $$ If not, would it be helpful if we increase the regularity of $\Gamma$ and $\Gamma_n$? i.e., change them to close $C^2$ curve?

$\endgroup$
  • $\begingroup$ Is this in 2D? Why do you say "the" solution, the boundary values on the curve are arbitrary? If you mean the solution of infimal Dirichlet integral, then in 2D you are looking at the conformal capacity, and in higher dimensions removing a curve makes no difference for the infimum. $\endgroup$ – user147263 Mar 15 '16 at 4:00
  • $\begingroup$ @404 yes, the problem is in 2D. The boundary values on the curve are free but I think it is true that the directional derivative on the curve is 0. $\endgroup$ – spatially Mar 15 '16 at 4:28
  • $\begingroup$ It would help to edit the question to clarify. $\endgroup$ – user147263 Mar 15 '16 at 4:34
  • $\begingroup$ This is true, but the proof is not easy. I would use conformal maps onto circular annuli $\{r_n<|z|<1\}$, where $r_n\to r$ by the doubly-connected version of the Caratheodory convergence theorem for conformal maps. The Dirichlet integral stays invariant, and so do the boundary conditions (although $\omega$ gets composed with the conformal map). One still has to consider the effect of this composition on the boundary value, and how it reflects in the integral... not fun. $\endgroup$ – user147263 Mar 15 '16 at 4:42
  • $\begingroup$ So, I would first thoroughly dig through the literature on capacity in the complex plane, looking for continuity results of this kind (sets converge in the Hausdorff metric). It's not continuous under Hausdorff convergence of general sets (e.g., finite sets can converge to an interval), but should be continuous for curves. $\endgroup$ – user147263 Mar 15 '16 at 4:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.